Effusion Gas Leakage: Estimating the Rate of Escape Through a Hole

[pentazoid]

Homework Statement

If you poke a hole in a container full of gas, the gass will start leaking out. In this problem, you weill make a rough estimate of the rate at which gas escapes through a hole.

If we now take away this small part of the wall of the container, the molecules that would have collidedwith it will instead escape through the hole. Assuming nothing enters through the hole , show that the number N of molecules inside the cotainer as a function of time is governed by the DE equation

dN/dt=-(A/2V)*sqrt(kT/m)*N

Homework Equations

Possible equations that pertain to this problem
P=-m*(delta(v₂)/delta(t))/A
v_x=sqrt(kT/m)
PV=Nmv_x^2

The Attempt at a Solution

PV=(-m*(delta(v_x)/delta(t))/A)*V=Nmv_x^2

N=((-m*(delta(v_x)/delta(t))/A)*V) /mv_x^2

m cancel and and I am left with :

N=((-(delta(v₂)/delta(t))/A)*V) /v_x^2

N=((-1/delta(t))/A)*V) /v_x^1= (-V*(delta(t))/A)/(v_x)

N/delta(t)= -V/A(v_x)=V/(A*sqrt(kT/m))

Something is wrong with my calculating because I think I am suppose to integrate N and my A nd V are in the wrong places.

 

[Count Iblis]

The flux of molecules is given by:

F = 1/4 n v (1)

where n is the number density and v the average velocity. This formula is always valid when the velocity dostribution is isotropic. In case of the Maxwell distribution, the average velocity is:

v = sqrt[8 k T/(pi m)] (2)

The answer should therefore be:

dN/dt = 1/4 N/V A sqrt[8 k T/(pi m)]

Clearly, the answer you were trying to derive is completely wrong in the first place.

You should study this topic from some decent statistical mechanics textbook. Formula (1) is not difficult to derive. It is a very important formula, so you should try to derive it. Formula (2) follows from a straightforward integration of Maxwell's velocity distribution.

 

[pentazoid]

The flux of molecules is given by:

F = 1/4 n v (1)

where n is the number density and v the average velocity. This formula is always valid when the velocity dostribution is isotropic. In case of the Maxwell distribution, the average velocity is:

v = sqrt[8 k T/(pi m)] (2)

The answer should therefore be:

dN/dt = 1/4 N/V A sqrt[8 k T/(pi m)]

Clearly, the answer you were trying to derive is completely wrong in the first place.

You should study this topic from some decent statistical mechanics textbook. Formula (1) is not difficult to derive. It is a very important formula, so you should try to derive it. Formula (2) follows from a straightforward integration of Maxwell's velocity distribution.

wait , how did you derived equations(1) and (2)? Your answer is wrong ; dN/dt=-V/A(vx)=V/(A*sqrt(kT/m))